How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$ How to prove this inequality?
If $a^{2}+b^{2}+c^{2}\leq 3$ and $a,b,c\in \Bbb R^+$, then
$$\left( a+b+c\right) \left( a+b+c-abc\right)\geq  2\left( a^{2}b+b^{2}c+c^{2}a\right) $$
I tried AM>GM but I couldn't get result
 A: Assume w.l.o.g that $a \leq b \leq c$.
Then we have : $(a+b+c)^3 \geq 7(a^2b+b^2c+c^2a)+6abc$; thus it is sufficient to prove: $(a+b+c)(a+b+c-abc) \geq \dfrac{2}{7}((a+b+c)^3-6abc)$.
Let $x=a+b+c$ and $y=abc$. The last inequality is then: $x^2-xy \geq \dfrac{2}{7}(x^3-6y)$, which is equivalent to: $x^2+\dfrac{12y}{7} \geq \dfrac{2x}{7}(x^2+\dfrac{7y}{2})$ and since $x \leq 3$ (using Cauchy-Schwarz on the original inequality), it is sufficient to prove: $x^2+\dfrac{12y}{7} \geq \dfrac{6}{7}(x^2+\dfrac{7y}{2})$, which is equivalent to: $x^2 \geq 9y$, which follows from A.M.-G.M. ($x \geq 3y^{1/3} \geq 3y^{1/2}$) where we used $y=abc\leq 1$.
A: Alternative solution:
Recall the known inequality $a^2b+b^2c+c^2a \le \frac{4}{27}(a+b+c)^3 - abc$ for $a, b, c \ge 0$.
See: https://artofproblemsolving.com/community/c6h478168p2677448, 
https://artofproblemsolving.com/community/c4h1705936p10979968, 
https://artofproblemsolving.com/community/c6h479531p2685016
It suffices to prove that
$(a+b+c)(a+b+c-abc) \ge \frac{8}{27}(a+b+c)^3 - 2abc$
or
$$(a+b+c)^2 + 2abc \ge \frac{8}{27}(a+b+c)^3 + (a+b+c)abc.$$
By using the method of @Aravind,  since $a+b+c \le 3\sqrt{\frac{a^2+b^2+c^2}{3}} \le 3$, it suffices to prove that
$$(a+b+c)^2 + 2abc \ge \frac{8}{27}(a+b+c)^2 \cdot 3 + 3abc$$
or
$$(a+b+c)^2 \ge 9abc.$$
Since $\frac{a+b+c}{3} \ge \sqrt[3]{abc}$ and $a+b+c\le 3$, we have $abc\le 1$ and $(a+b+c)^2 \ge 9(abc)^{2/3} \ge 9abc$. 
We are done.
A: Another way.
We need to prove that
$$(a+b+c)^2\geq2\sum_{cyc}a^2b+abc(a+b+c)$$ and since $$\frac{3}{a^2+b^2+c^2}\geq1$$ and $$1\geq\frac{a^2+b^2+c^2}{3},$$ it's enough to prove that:
$$(a+b+c)^2\sqrt{\frac{a^2+b^2+c^2}{3}}\geq2\sum_{cyc}a^2b+\sqrt{\frac{3}{a^2+b^2+c^2}}abc(a+b+c)$$ or
$$(a+b+c)^2(a^2+b^2+c^2)\geq2\sum_{cyc}a^2b\sqrt{3(a^2+b^2+c^2)}+3abc(a+b+c).$$ 
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, $abc=w^3$ and $u=tv$.
Thus, $t\geq1$ and since by Rearrangement $\sum_{cyc}a^2b\leq4u^3-w^3$, it's enough to prove that:
$$9u^2(9u^2-6v^2)\geq2(4u^3-w^3)\sqrt{3(9u^2-6v^2)}+9uw^3$$ or
$$9u^2(3u^2-2v^2)-8u^3\sqrt{3u^2-2v^2}\geq(3u-2\sqrt{3u^2-2v^2})w^3.$$
Now, if $$(3u-2\sqrt{2u^2-2v^2}<0$$ so our inequality is obviously true because
$$9u^2(3u^2-2v^2)-8u^3\sqrt{3u^2-2v^2}=$$
$$=u^2\sqrt{3u^2-2v^2}(9\sqrt{3u^2-2v^2}-8u)\geq u^2\sqrt{3u^2-2v^2}(9\sqrt{u^2}-8u)>0.$$
But for $3u-2\sqrt{2u^2-2v^2}\geq0$ since $w^3\leq v^3,$ it's enough to prove that
$$9t^2(3t^2-2)-8t^3\sqrt{3t^2-2}\geq3t-2\sqrt{3t^2-2}$$ or
$$27t^4-18t^2-3t\geq(8t^3-2)\sqrt{3t^2-2}$$ or
$$(t-1)(537t^7+537t^6-307t^5-373t^4-49t^3-5t^2-8t-8)\geq0,$$ which is obvious for $t\geq1$.
